Question

If \(A,B\) are the feet of the perpendiculars drawn from \(P(3,1)\) on the pair of lines \[ 2x^2+3xy-2y^2+3x+y+1=0 \] and \(Q\) is the point of intersection of the pair of lines, then the area of the quadrilateral \(PAQB\) is

• A. \(25\)
• B. \(\dfrac{36}{5}\)
• C. \(\dfrac{25}{6}\)
• D. \(\dfrac{49}{8}\) \bigskip

Hint:

Whenever a quadratic equation represents a pair of straight lines, factorize it first. If the two lines turn out to be perpendicular, the geometry becomes much simpler and area calculations often reduce to products of perpendicular distances.

Answer 1

The Correct Option is B

Concept: Factor the second-degree equation into lines L1 and L2. The area of quadrilateral PAQB, where P is a point and Q is the intersection of L1, L2, can be found as sum of areas of two triangles. Use perpendicular distance from P to the lines and the fact that L1 ⟂ L2.

Step 1: Factor: \(2x^2+3xy-2y^2+3x+y+1 = (2x-y+1)(x+2y+1)=0\). L1: \(2x-y+1=0\), L2: \(x+2y+1=0\). Intersection Q: solving gives \(x=-3/5, y=-1/5\). P is (3,1).

Step 2: Distance from P to L1: \(d_1 = \frac{|2(3)-1+1|}{\sqrt5} = \frac{6}{\sqrt5}\). To L2: \(d_2 = \frac{|3+2+1|}{\sqrt5} = \frac{6}{\sqrt5}\).

Step 3: Check angle between L1 and L2. Slopes: 2 and -1/2, product = -1, so they are perpendicular.

Step 4: The quadrilateral PAQB consists of two right triangles with legs PA, AQ and PB, BQ? Actually, P to Q and the feet to the lines. Since lines are perpendicular, the area simplifies to \(d_1 \cdot d_2 = \frac{36}{5}\).

Step 5: Write the final answer. \(\boxed{\frac{36}{5}}\)